Problem: A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.

Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).

Output

Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.

Examples

Input

1 0


Output

1 2

Input

2 1


Output

3 2 1 4


Input

4 0


Output

2 7 4 6 1 3 5 8

Note

Record |x| represents the absolute value of number x. 

In the first sample |1 - 2| - |1 - 2| = 0.

In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.

In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.

Solution (in JAVA):

Solution:

import java.io.*;
public class B359 
{
    public static void main(String args[])throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String s[]=br.readLine().split(" ");
        int n=Integer.parseInt(s[0]);
        int k=Integer.parseInt(s[1]);
        if(k==0)
        {
            for(int i=1;i<=2*n;i++)
            {
                System.out.print(i+" ");
            }
        }
        else
        {
            int c=0,flag=0;
            int i=1,j=i+k;
            int cou[]=new int[2*n+1];
            while(c<2)
            {
                if(flag==0)
                {
                    System.out.print(i+" "+j+" ");
                    i++;
                    j++;
                    flag=1;
                    cou[i-1]=1;
                    cou[j-1]=1;
                    c++;
                }
                else
                {
                    if(cou[i]==0 && cou[j]==0)
                    {
                        System.out.print(j+" "+i+" ");
                        cou[i]=1;
                        cou[j]=1;
                        i++;
                        j++;
                        flag=0;
                        c++;
                    }
                    else
                    {
                        i++;
                        j++;
                    }
                }
            }
            i=3;
            j=4;
            int z=c;
            while(z<n)
            {
                if(cou[i]==1)
                    i++;
                if(cou[j]==1)
                    j++;
                if(i==j)
                    j++;
                if(cou[i]==0 && cou[j]==0)
                {
                    cou[i]=1;
                    cou[j]=1;
                    System.out.print(i+" "+j+" ");
                    i++;
                    j++;
                    z++;
                }
            }
        }
    }
}